Question:
If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots$ upto 100 terms, then $\alpha$ is :
Correct Option: 1
Solution:
$\operatorname{Cot}^{-1}(\alpha)=\cot ^{-1}(2)+\cot ^{-1}(8)+\cot ^{-1}(18)+\ldots \ldots$
$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{2}{4 n^{2}}\right)$
$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\right)$
$=\sum_{n=1}^{100} \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$
$=\tan ^{-1} 201-\tan ^{-1} 1$
$=\tan ^{-1}\left(\frac{200}{202}\right)$
$\therefore \cot ^{-1}(\alpha)=\cot ^{-1}\left(\frac{202}{200}\right)$
$\alpha=1.01$
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